I am trying to show that $x^2+x+1$ is irreducible over $GF(2^n)$ for each odd $n$.
Clearly it's true for $n=1$ since then $GF(2) = \mathbb{Z}_2$.
I think the idea is to assume $u$ is a root of $x^2+x+1$ and then calculate $[GF(p^n)(u):GF(p^n)]$. But I am having trouble doing that.
Observe that if $x^2+x+1$ has roots in a field $F$, then they are cube roots of unity, i.e. they are elements of order $3$ in the multiplicative group $F^\times$. Also recall that $\mathbb{F}_{p^n}^\times$ is a cyclic group of size $p^n-1$.
When does the cyclic group $\mathbb{F}_{2^n}^\times$ have an element of order $3$?